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\begin{document}
\maketitle
\section{spline}
[a, b]进行分割，$a = x_0 < x_1 < x_2 < ... < x_N = b$, $s(x) \in C^k[a,b]$,s(x)在子区间$[x_i, x_{i+1}]$上的限制：$s(x)|_{[x_i, x_{i+1}]}$,是一个n次多项式。$s(x) \in \mathbb{S}^k_n$.
\subsection{条件对比}
$s(x)|_{[x_i,x_{i+1}]} = p_i(x)$
需要$(n+1)*(N-1)$个条件，已知$s^{(d)}|_{[x_{i-1}],x_i} = s^{(d)}|_{[x_i,x_{i+1}]}, d = 0, 1, ...,k ; s(x_i) = f(x_i), i = 0, ..., N$\\
$k = n-1$, 还需要两个条件。

\subsection{$\mathbb{S}^3_2$}
$m_i = f'(x_i), K = f[x_i, x_{i+1}]$\\
$\lambda_i*m_{i-1} + 2*m_{i} + \mu_i*m_{i+1} = 3*\mu_i*f[x_i,x_{i+1}] + 3*\lambda_i*f[x_{i-1}, x_i], i = 2,3,...,N-1$\\
$\lambda_i = \dfrac{x_{i+1}- x_i}{x_{i+1}-x_{i-1}}, \mu_i =\dfrac{x_{i}- x_{i-1}}{x_{i+1}-x_{i-1}}$，已知。$m_i$未知。\\
\begin{align}
  \lambda_2*m_1 + 2*m_2 + \mu_2*m_3 = &b_2 \\
  \lambda_3*m_2 + 2*m_3 + &\mu_3*m_4 = b_3 \\
  \lambda_4*m_3 + &2*m_4 + \mu_4*m_5 = b_4 \\
  ...\\
  &\lambda_{N-1}*m_{N-2} + 2*m_{N-1} + \mu_{N-1}*m_N = b_{N-1} \\
\end{align}
N-2个方程，$m_i, i = 1,2,...,N$，N个未知量。\\
$m_1 = f'(x_1) = f'(a)$\\
$m_N = f'(x_N) = f'(b)$
\begin{align}
  2*m_2 + \mu_2*m_3 = &b_2 - \lambda_2*m_1 \\
  \lambda_3*m_2 + 2*m_3 + &\mu_3*m_4 = b_3 \\
  \lambda_4*m_3 + &2*m_4 + \mu_4*m_5 = b_4 \\
  ...\\
  &\lambda_{N-1}*m_{N-2} + 2*m_{N-1} = b_{N-1} - \mu_{N-1}*m_N \\
\end{align}
\begin{align}
  2 + \mu_2 = &b_2 - \lambda_2*m_1 \\
  \lambda_3 + 2 + &\mu_3 = b_3 \\
  \lambda_4 + &2 + \mu_4 = b_4 \\
  ...\\
  &\lambda_{N-1} + 2 = b_{N-1} - \mu_{N-1}*m_N \\
\end{align}
三斜率方程组，用$M_i$代替$m_i$得到类似的三弯距方程组。

\subsection{边界条件}
\subsubsection{complete,D1完全样条}
$m_1 = f'(x_1) = f'(a)$\\
$m_N = f'(x_N) = f'(b)$

\subsubsection{D2}
在端点给定二阶导数的样条。\\
和3.7搭配比较自然。

\subsubsection{自然样条}
$M_1 = f''(a) = M_N = f''(b) = 0$\\
端点自然放松不加约束。

\subsubsection{}
\end{document}